Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $z = \dfrac{3p - 24}{3p - 30} \div \dfrac{p^2 - 3p - 40}{4p^2 - 60p + 200} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{3p - 24}{3p - 30} \times \dfrac{4p^2 - 60p + 200}{p^2 - 3p - 40} $ First factor out any common factors. $z = \dfrac{3(p - 8)}{3(p - 10)} \times \dfrac{4(p^2 - 15p + 50)}{p^2 - 3p - 40} $ Then factor the quadratic expressions. $z = \dfrac {3(p - 8)} {3(p - 10)} \times \dfrac {4(p - 10)(p - 5)} {(p - 8)(p + 5)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {3(p - 8) \times 4(p - 10)(p - 5) } {3(p - 10) \times (p - 8)(p + 5) } $ $z = \dfrac {12(p - 10)(p - 5)(p - 8)} {3(p - 8)(p + 5)(p - 10)} $ Notice that $(p - 8)$ and $(p - 10)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {12(p - 10)(p - 5)\cancel{(p - 8)}} {3\cancel{(p - 8)}(p + 5)(p - 10)} $ We are dividing by $p - 8$ , so $p - 8 \neq 0$ Therefore, $p \neq 8$ $z = \dfrac {12\cancel{(p - 10)}(p - 5)\cancel{(p - 8)}} {3\cancel{(p - 8)}(p + 5)\cancel{(p - 10)}} $ We are dividing by $p - 10$ , so $p - 10 \neq 0$ Therefore, $p \neq 10$ $z = \dfrac {12(p - 5)} {3(p + 5)} $ $ z = \dfrac{4(p - 5)}{p + 5}; p \neq 8; p \neq 10 $